Energy with Non-Conservative Forces: Videos & Practice Problems

Incorporating nonconservative forces like applied forces and friction into the conservation of energy equation reveals that mechanical energy is not conserved when work is done on a system. For example, when a hockey puck is pushed, the work done increases its kinetic energy. The full conservation of energy equation is expressed as ${\frac{1}{2}}^{\mathrm{mv}}+W={\frac{1}{2}}^{\mathrm{mv}}$, where W represents work done by nonconservative forces. This understanding is crucial for analyzing energy transformations in various physical scenarios.

Energy Conservation with Non-Conservative Forces

In the study of mechanics, the conservation of energy principle is crucial for understanding how energy is transferred and transformed within a system. Traditionally, this principle applies to conservative forces, such as gravity and springs, where the total mechanical energy remains constant. However, when nonconservative forces, like applied forces and friction, come into play, the situation changes significantly.

Consider a practical example involving a hockey puck. Initially, the puck is moving with a speed of 4 m/s. When a player applies a force of 200 N to push the puck over a distance of 0.3 meters, the puck's speed increases due to the work done on it. To analyze this scenario, we can use the conservation of energy equation, which incorporates both kinetic energy and the work done by nonconservative forces.

The general form of the conservation of energy equation is:

\[ K + U_{\text{initial}} + W_{\text{nc}} = K + U_{\text{final}} \]

In this case, since the puck is moving along a horizontal surface, the potential energy (U) remains constant (U_initial = U_final = 0). Therefore, the equation simplifies to:

\[ K_{\text{initial}} + W_{\text{nc}} = K_{\text{final}} \]

Here, \(K\) represents kinetic energy, calculated using the formula:

\[ K = \frac{1}{2} m v^2 \]

Where \(m\) is the mass of the puck and \(v\) is its velocity. The work done by nonconservative forces (\(W_{\text{nc}}\)) is the sum of the work done by applied forces and friction. In this scenario, since the puck is on smooth ice, we only consider the work done by the applied force:

\[ W_{\text{nc}} = F_a \cdot d \cdot \cos(\theta) \]

Given that the applied force and the distance are in the same direction, the angle \(\theta\) is 0 degrees, making \(\cos(0) = 1\). Thus, the work done by the applied force is:

\[ W_{\text{nc}} = 200 \, \text{N} \cdot 0.3 \, \text{m} = 60 \, \text{J} \]

Now, substituting the values into the energy equation, we can find the final kinetic energy:

\[ \frac{1}{2} (0.5 \, \text{kg}) (4 \, \text{m/s})^2 + 60 \, \text{J} = \frac{1}{2} (0.5 \, \text{kg}) v_{\text{final}}^2 \]

Calculating the initial kinetic energy gives:

\[ \frac{1}{2} (0.5) (16) = 4 \, \text{J} \]

Thus, the equation becomes:

\[ 4 \, \text{J} + 60 \, \text{J} = \frac{1}{2} (0.5) v_{\text{final}}^2 \]

Solving for \(v_{\text{final}}\), we find:

\[ 64 \, \text{J} = \frac{1}{2} (0.5) v_{\text{final}}^2 \]

Rearranging gives:

\[ v_{\text{final}}^2 = \frac{64 \times 2}{0.5} = 256 \]

Taking the square root results in:

\[ v_{\text{final}} = 16 \, \text{m/s} \]

This example illustrates how the work done by nonconservative forces affects the total mechanical energy of the system. The initial kinetic energy of 4 J, combined with the 60 J of work done, results in a final kinetic energy of 64 J, confirming that mechanical energy is not conserved when nonconservative forces are present. Understanding these principles is essential for solving problems involving energy transformations in various physical contexts.

Box Sliding on Rough Patch

In this scenario, we analyze a block of unknown mass sliding on a surface with an initial velocity of \( v_i = 30 \, \text{m/s} \). Upon encountering a rough patch, the block eventually comes to a stop, and we aim to calculate the distance \( d \) it travels before stopping. To solve this, we apply the principle of conservation of energy, focusing on the initial and final states of the block.

The conservation of energy equation can be expressed as:

\[ K_{\text{initial}} + U_{\text{initial}} + W_{\text{non-conservative}} = K_{\text{final}} + U_{\text{final}} \]

In this case, the initial kinetic energy \( K_{\text{initial}} \) is given by:

\[ K_{\text{initial}} = \frac{1}{2} m v_i^2 \]

Since there is no gravitational potential energy or spring energy involved, both \( U_{\text{initial}} \) and \( U_{\text{final}} \) are zero. The final kinetic energy \( K_{\text{final}} \) is also zero because the block comes to a stop. The work done by non-conservative forces, primarily friction, is represented as \( W_{\text{non-conservative}} = -f_k d \), where \( f_k \) is the force of kinetic friction.

The force of kinetic friction can be expressed as:

\[ f_k = \mu_k N \]

For a block on a flat surface, the normal force \( N \) equals the weight of the block, \( mg \). Thus, we have:

\[ f_k = \mu_k mg \]

Substituting this into our work equation gives:

\[ \frac{1}{2} m v_i^2 - \mu_k mg d = 0 \end{align*} \]

Rearranging this equation leads to:

\[ \frac{1}{2} m v_i^2 = \mu_k mg d \end{align*} \]

Notably, the mass \( m \) cancels out, allowing us to solve for the distance \( d \) without knowing the mass:

\[ d = \frac{v_i^2}{2 \mu_k g} \end{align*} \]

Substituting the known values, where \( v_i = 30 \, \text{m/s} \), \( \mu_k = 0.6 \), and \( g = 9.8 \, \text{m/s}^2 \), we calculate:

\[ d = \frac{30^2}{2 \times 0.6 \times 9.8} \approx 76.5 \, \text{meters} \end{align*} \]

This result illustrates how the initial kinetic energy of the block is dissipated by the work done against friction, ultimately bringing the block to a stop. The work done by non-conservative forces, such as friction, accounts for the energy lost in the system, demonstrating the interplay between kinetic energy and frictional forces in motion.

An 800kg car skids to a stop from 30m/s through a distance of 90m. Calculate the coefficient of friction between the car and the road.

• A. 0.51
• B. 1.23
• C. 0.73
• D. 0.02

Answer: (A)

Energy Conservation with Air Resistance

In this scenario, we analyze the motion of a 2-kilogram object dropped from a height of 80 meters, reaching the ground with a final speed of 30 m/s while experiencing air resistance. To understand the work done by air resistance, we apply the principle of conservation of energy, which states that the total mechanical energy in a system remains constant if only conservative forces are acting. However, in this case, air resistance is a non-conservative force that does work on the object.

The conservation of energy equation can be expressed as:

\[ K_{\text{initial}} + U_{\text{initial}} + W_{\text{nonconservative}} = K_{\text{final}} + U_{\text{final}} \]

Here, the initial kinetic energy \(K_{\text{initial}}\) is zero since the object starts from rest, and the initial gravitational potential energy \(U_{\text{initial}}\) is given by \(mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^2\)), and \(h\) is the height (80 m). The final potential energy \(U_{\text{final}}\) is zero at ground level.

Substituting the known values, we have:

\[ U_{\text{initial}} = mgh = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 80 \, \text{m} = 1568 \, \text{J} \]

The final kinetic energy \(K_{\text{final}}\) is given by:

\[ K_{\text{final}} = \frac{1}{2} mv^2 = \frac{1}{2} \times 2 \, \text{kg} \times (30 \, \text{m/s})^2 = 900 \, \text{J} \]

Incorporating these into the conservation of energy equation, we can solve for the work done by air resistance \(W_{\text{air}}\):

\[ 1568 \, \text{J} + W_{\text{air}} = 900 \, \text{J} \]

Rearranging gives:

\[ W_{\text{air}} = 900 \, \text{J} - 1568 \, \text{J} = -668 \, \text{J} \]

The negative sign indicates that air resistance does work against the motion of the object, removing energy from the system.

Next, to find the average force of air resistance, we can use the work-energy principle, which states that work done by a constant force is equal to the force multiplied by the distance over which it acts:

\[ W = F_{\text{air}} \cdot d \cdot \cos(\theta) \

In this case, the angle \(\theta\) between the force of air resistance (acting upwards) and the displacement (downwards) is 180 degrees, making \(\cos(180^\circ) = -1\). Thus, we can simplify the equation to:

\[ W_{\text{air}} = F_{\text{air}} \cdot d \]

Substituting the known values, where \(d = -80 \, \text{m}\) (the distance fallen), we have:

\[ -668 \, \text{J} = F_{\text{air}} \cdot (-80 \, \text{m}) \]

Solving for \(F_{\text{air}}\) gives:

\[ F_{\text{air}} = \frac{668 \, \text{J}}{80 \, \text{m}} = 8.35 \, \text{N} \]

This value represents the average force of air resistance acting on the object during its fall. Understanding these concepts allows us to analyze the effects of non-conservative forces like air resistance in mechanical systems.

Here’s what students ask on this topic:

What is the difference between conservative and nonconservative forces?

Conservative forces, such as gravity and spring forces, are path-independent and do not dissipate mechanical energy. The work done by these forces depends only on the initial and final positions, not the path taken. Nonconservative forces, like friction and applied forces, are path-dependent and can dissipate mechanical energy as heat or other forms. The work done by nonconservative forces can add or remove energy from a system, making the total mechanical energy not conserved.

How do you incorporate nonconservative forces into the conservation of energy equation?

To incorporate nonconservative forces into the conservation of energy equation, you add a term for the work done by these forces. The full conservation of energy equation is:

$\frac{1}{2}m{v}_{\mathrm{initial}}+W=\frac{1}{2}m{v}_{\mathrm{final}}$

where $W$ represents the work done by nonconservative forces. This term accounts for any energy added or removed from the system by forces like friction or applied forces.

How do you calculate the work done by nonconservative forces?

The work done by nonconservative forces is calculated using the formula:

$W=Fdcos\theta$

where $F$ is the force applied, $d$ is the distance over which the force is applied, and $\theta$ is the angle between the force and the direction of motion. This formula accounts for the component of the force that does work in the direction of the displacement.

Why is mechanical energy not conserved when nonconservative forces are present?

Mechanical energy is not conserved when nonconservative forces are present because these forces can add or remove energy from the system. For example, friction converts mechanical energy into heat, and applied forces can increase or decrease the system's energy. This results in a change in the total mechanical energy, making it different from the initial mechanical energy. The work done by nonconservative forces accounts for this energy change.

Can you provide an example of a problem involving nonconservative forces and how to solve it?

Consider a hockey puck of mass 0.5 kg initially moving at 4 m/s. An applied force of 200 N pushes it over a distance of 0.3 m. To find the final speed, use the full conservation of energy equation:

$\frac{1}{2}m{v}_{\mathrm{initial}}+W=\frac{1}{2}m{v}_{\mathrm{final}}$

Calculate the work done by the applied force:

$W=Fdcos0=200\times 0.3\times 1=60J$

Substitute values into the energy equation:

$\frac{1}{2}\left(0.5\right)\left(4{)}^2\right)+60=\frac{1}{2}\left(0.5\right)v_{\mathrm{final}}{)}^2$

Solve for $v_{\mathrm{final}}$ to get 16 m/s.